= Solution
The elemental <torque> about the origin is $d\boldsymbol\tau=\mathbf r\times d\mathbf F$. Substitution of part (a) gives
$$
d\boldsymbol\tau=Ia^2\bigl(-B_y\sin^2\theta-B_x\sin\theta\cos\theta,\ B_y\sin\theta\cos\theta+B_x\cos^2\theta,\ 0\bigr)\,d\theta.
$$
Only its $x$ component drives the allowed rotation about the $x$ axis. Using $\int_0^{2\pi}\sin^2\theta\,d\theta=\pi$ and $\int_0^{2\pi}\sin\theta\cos\theta\,d\theta=0$, the projected <torque> is
$$
\boxed{\boldsymbol\tau_{\text{about }x}=-I\pi a^2B_y\hat{\mathbf x}.}
$$
The <magnetic dipole moment> is $\mathbf m=I\pi a^2\hat{\mathbf z}$, so the full electromagnetic <torque> is
$$
\mathbf m\times\mathbf B=I\pi a^2(-B_y\hat{\mathbf x}+B_x\hat{\mathbf y}).
$$
Its $x$ component agrees exactly. The $y$ component is a real electromagnetic <torque>, but the bearing constraining motion to the $x$ axis supplies a cancelling reaction <torque>. For an allowed infinitesimal rotation $d\boldsymbol\alpha=\hat{\mathbf x}\,d\alpha$, only $\boldsymbol\tau\cdot d\boldsymbol\alpha=\tau_xd\alpha$ does mechanical <work>.
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